(5-z)(7z+4)=62

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Solution for (5-z)(7z+4)=62 equation: 



(5-z)(7z+4)=62

We move all terms to the left:

(5-z)(7z+4)-(62)=0

We add all the numbers together, and all the variables

(-1z+5)(7z+4)-62=0

We multiply parentheses ..

(-7z^2-4z+35z+20)-62=0

We get rid of parentheses

-7z^2-4z+35z+20-62=0

We add all the numbers together, and all the variables

-7z^2+31z-42=0

a = -7; b = 31; c = -42;
Δ = b2-4ac
Δ = 312-4·(-7)·(-42)
Δ = -215
Delta is less than zero, so there is no solution for the equation

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